281. If   2x −	1	= 5, x ≠ 0, then the value of		x2 +	1	− 2		is :
     	2x				16x2

1. 1. 19

      4

   
   
   

   2. 23

      4

   
   
   

   3. 27

      4

   
   
   

   4. 31

      4

   
   
   

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1. View Hint View Answer Discuss in Forum

   2x -	1	= 5
   	2x

   
   On dividing by 2,
   

   x -	1	=	5
   	4x		2

   
   On squaring both sides
   

   	x -	1		²	=		5		²	=	25
   		4x					2				4

   
   

   ⇒ x² +	1	-2 × x ×	1	=	25
   	16x²		4x		4

   
   

   ⇒ x² +	1	=	25	+	1
   	16x²		4		2

   
   

   =	25 + 2	=	27
   	4		4

   
   

   ⇒ x² +	1	- 2 =	27	- 2 =	27 - 8	=	19
   	16x²		4		4		4

   Correct Option: A

   2x -	1	= 5
   	2x

   
   On dividing by 2,
   

   x -	1	=	5
   	4x		2

   
   On squaring both sides
   

   	x -	1		²	=		5		²	=	25
   		4x					2				4

   
   

   ⇒ x² +	1	-2 × x ×	1	=	25
   	16x²		4x		4

   
   

   ⇒ x² +	1	=	25	+	1
   	16x²		4		2

   
   

   =	25 + 2	=	27
   	4		4

   
   

   ⇒ x² +	1	- 2 =	27	- 2 =	27 - 8	=	19
   	16x²		4		4		4

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282. If   a +	1	= 1, b +	1	= 1, then the value of (abc) is :
     	b		c

1. 1. 0

   
   
   

   2. –1

   
   
   

   3. 1

   
   
   

   4. ab

   
   
   

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1. View Hint View Answer Discuss in Forum

   a +	1	= 1
   	a

   
   

   ⇒ a = 1 -	1	=	b - 1
   	b		b

   
   Again,
   

   b +	1	= 1
   	c

   
   

   ⇒ b = 1 -	1	=	c - 1
   	c		c

   
   

   ∴ a =			c - 1	- 1
   	b - 1	=	c
   	b		c - 1
   			c

   
   

   =	c - 1 - c	=	- 1
   	c - 1		c - 1

   
   

   ∴ abc =	- 1	×	c - 1	× c = - 1
   	c - 1		c

   Correct Option: B

   a +	1	= 1
   	a

   
   

   ⇒ a = 1 -	1	=	b - 1
   	b		b

   
   Again,
   

   b +	1	= 1
   	c

   
   

   ⇒ b = 1 -	1	=	c - 1
   	c		c

   
   

   ∴ a =			c - 1	- 1
   	b - 1	=	c
   	b		c - 1
   			c

   
   

   =	c - 1 - c	=	- 1
   	c - 1		c - 1

   
   

   ∴ abc =	- 1	×	c - 1	× c = - 1
   	c - 1		c

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283. If	3	=	a	+	b	be an identify, then the value of b is :
     	(x + 2)(2x + 1)		2x + 1		x + 2

1. 1. 0

   
   
   

   2. –1

   
   
   

   3. 2

   
   
   

   4. 3

   
   
   

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1. View Hint View Answer Discuss in Forum

   	3	=	a	+	b
   	(x + 2)(2x + 1)		2x + 1		x + 2

   
   

   ⇒	3	=	a(x + 2) + b (2x + 1)
   	(x + 2)(2x + 1)		(2x + 1)(x + 2)

   
   ⇒ 3 = ax + 2a + 2bx + b
   ⇒ 3 = ax + 2bx + 2a + b
   ⇒ 3 = x (a + 2b) + (2a + b)
   On comparing the respective co– efficients,
   a + 2b = 0
   ⇒ a = – 2b ..... (i)
   and, 2a + b = 3 2 (– 2b) + b = 3
   ⇒ – 4b + b = 3
   

   ⇒ – 3b = 3 ⇒ b =	- 3	= - 1
   	3

   Correct Option: B

   	3	=	a	+	b
   	(x + 2)(2x + 1)		2x + 1		x + 2

   
   

   ⇒	3	=	a(x + 2) + b (2x + 1)
   	(x + 2)(2x + 1)		(2x + 1)(x + 2)

   
   ⇒ 3 = ax + 2a + 2bx + b
   ⇒ 3 = ax + 2bx + 2a + b
   ⇒ 3 = x (a + 2b) + (2a + b)
   On comparing the respective co– efficients,
   a + 2b = 0
   ⇒ a = – 2b ..... (i)
   and, 2a + b = 3 2 (– 2b) + b = 3
   ⇒ – 4b + b = 3
   

   ⇒ – 3b = 3 ⇒ b =	- 3	= - 1
   	3

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284. If		x +	1		= 5, then find the value of	6x
     			x			x2 + x + 1

1. 1. 3

   
   
   

   2. 2

   
   
   

   3. 1

   
   
   

   4. 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   It is given,   x +	1	= 5
   	x

   
   

   Expression =	6x
   	x2 + x + 1

   
   

   =	6x					=	6
   	x
   			x + 1 +	1				x +	1	+ 1
   				x					x

   
   

   =	6	=	6	= 1
   	5 + 1		6

   Correct Option: C

   It is given,   x +	1	= 5
   	x

   
   

   Expression =	6x
   	x2 + x + 1

   
   

   =	6x					=	6
   	x
   			x + 1 +	1				x +	1	+ 1
   				x					x

   
   

   =	6	=	6	= 1
   	5 + 1		6

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285. If a + b = 5 and a – b = 3, then the value of (a² + b²) is :

1. 1. 17

   
   
   

   2. 18

   
   
   

   3. 19

   
   
   

   4. 20

   
   
   

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1. View Hint View Answer Discuss in Forum

   a + b = 5
   a – b = 3
   ∵  (a + b)² + (a – b)²
   = 2 (a² + b²)
   ⇒  2 (a² + b²) = 5² + 3²
   = 25 + 9 = 34
   

   ⇒  a2 + b2 =	34	= 17
   	2

   Correct Option: A

   a + b = 5
   a – b = 3
   ∵  (a + b)² + (a – b)²
   = 2 (a² + b²)
   ⇒  2 (a² + b²) = 5² + 3²
   = 25 + 9 = 34
   

   ⇒  a2 + b2 =	34	= 17
   	2

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